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Between any 2 distinct real numbers, does there exist a real number such that its decimal expansion terminates in base 10?

Also, does this result hold in any natural number base? Thank you.

Arturo Magidin
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Brothersquid
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  • A real number has a decimal expansion that terminates in base $10$ if and only if it is a rational number and its denominator, when written in least terms, is of the form $2^a5^b$ with $a,b$ natural numbers. For base $b$, the expansion terminates if and only if all prime factors of the denominator divide $b$. – Arturo Magidin Dec 09 '22 at 20:07
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    This isn't difficult to prove, however ... what have you tried? – rtybase Dec 09 '22 at 20:14
  • I was thinking of taking of taking $a,b\in \mathbb{R}, 0 < a <b$ and considering the sequence of finite decimal expansions for $a$ and $b$, $a_n$ and $b_n$ respectively, which each converge to $a$ and $b$. Since $a \neq b$, I think that $a_n$ and $b_n$ need to disagree at some finite position $i$. $a_i$ should then be the desired construction. – Brothersquid Dec 09 '22 at 20:24
  • @Brothersquid You should have written that in your question. My answer below is essentially a waste of time if you knew that already. – preferred_anon Dec 09 '22 at 20:27
  • Well, I appreciated the input nonetheless. – Brothersquid Dec 09 '22 at 20:32

1 Answers1

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Of course. Imagine being given two infinite decimals (or any base $b$ expansion) and trying to prove that one is bigger than the other.

  • First you look at the digits in the most significant position. If one is bigger than the other (where missing digits count as leading 0s), then that's the bigger number.
  • Otherwise, you look at the second most significant digits. Are they different? If so, you have a bigger number. If not...
  • continue forever. If the numbers are different, you will eventually find a position where they differ.

If you understand this procedure, it is obvious that there is a terminating decimal that is at most as big as the bigger number. Can you prove the rest?

preferred_anon
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